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3 years ago

Charlie runs at an average speed of 6.5 km/hr. If he runs for 1.5 hours, how far has he traveled?

Physics

1 answer:

VikaD [51]3 years ago

8 0

Answer:

9.75 km

Explanation:

Charlie runs 6.5 km/hr

-> Charlie wants to run for 1.5 hours

6.5km + 6.5km/2

= 6.5 km + 3.25km

= 9.75 km

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it is a science question and the question is that which type of cloud would most likely produce a thunderstorm

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A cumulonimbus cloud most likely would.

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4 years ago

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Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

Romashka [77]

Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )

$x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)$

$x_{C} = 128.995\,m$

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

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3 years ago

How long does it take (in minutes) for light to reach mercury from the sun, a distance of 6.02 × 107 km?

Ksju [112]

Light travels at a speed of:
$c=3 \cdot 10^8 m/s$
The distance between Mercury and Sun is $S=6.02 \cdot 10^7 km=6.02 \cdot 10^{10} m$ , so the time it takes is
$t= \frac{S}{c}= \frac{6.02 \cdot 10^{10} m}{3 \cdot 10^8 m/s} =200.7 s$
if we want to convert this into minutes, keeping in mind that 1 min = 60 seconds, we should divide this value by 60:
$t= \frac{200.7 s}{60 s/min}=3.34 min$

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When a body strictly moves on a curve, it's velocity at a point is tangential to the curve at that point.

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In modern fixtures, neutral wire is white and the hot wire is red or black. In some types of fixtures, both wires will be of the same color.

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